Problem: The polar curve $r(\theta)=1-2\theta$ is graphed for $0\leq\theta\leq\pi$. Let $R$ be the region in the third quadrant enclosed by the curve and the $y$ -axis, as shown in the graph. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{\scriptsize\dfrac{\pi}{4}}^{\scriptsize\dfrac{\pi}{2}}\left( \dfrac{1}{4}-\theta+\theta^2\right)d\theta$ (Choice B) B $ \int_{\scriptsize\dfrac{\pi}{4}}^{\scriptsize\dfrac{\pi}{2}}\left( \dfrac{1}{2}-2\theta+2\theta^2\right)d\theta$ (Choice C) C $ \int_{\scriptsize\dfrac12}^{\scriptsize\dfrac{\pi}{2}}\left( \dfrac{1}{4}-\theta+\theta^2\right)d\theta$ (Choice D) D $ \int_{\scriptsize\dfrac12}^{\scriptsize\dfrac{\pi}{2}}\left( \dfrac{1}{2}-2\theta+2\theta^2\right)d\theta$
Solution: This is the formula for the area enclosed by a polar curve $r(\theta)$ between $\theta=\alpha$ and $\theta=\beta$ : $ \int_{\alpha}^{\beta}\dfrac{1}{2}\left(r(\theta)\right)^{2}d\theta$ We know $r(\theta)$ but we still need to figure out $\alpha$ and $\beta$. Notice that the curve goes through the third and fourth quadrants even though it's graphed for $0\leq\theta\leq\pi$. That's because after a certain $\theta$ -value, the sign of $r(\theta)$ is negative. $\alpha$ is the first (and only) non-negative $\theta$ -value for which $r(\theta)=0$. So $\alpha=\dfrac12$. $\beta$ is the first $\theta$ -value after $\alpha$ where the curve intersects the $y$ -axis. So $\beta=\dfrac{\pi}{2}$. Let's plug ${r(\theta)=1-2\theta}$, ${\alpha=\dfrac12}$, and ${\beta=\dfrac{\pi}{2}}$ into the formula and expand the parentheses: $\begin{aligned} &\phantom{=} \int_{\alpha}^{\beta}\dfrac{1}{2}\left({r(\theta)}\right)^{2}d\theta \\\\ &= \int_{{\scriptsize\dfrac12}}^{{\scriptsize\dfrac{\pi}{2}}}\dfrac{1}{2}\left({1-2\theta}\right)^{2}d\theta \\\\ &= \int_{\scriptsize\dfrac12}^{\scriptsize\dfrac{\pi}{2}}\dfrac{1}{2}\left( 1-4\theta+4\theta^2\right)d\theta \\\\ &= \int_{\scriptsize\dfrac12}^{\scriptsize\dfrac{\pi}{2}}\left( \dfrac{1}{2}-2\theta+2\theta^2\right)d\theta \end{aligned}$ In conclusion, this integral represents the area of region $R$ : $ \int_{\scriptsize\dfrac12}^{\scriptsize\dfrac{\pi}{2}}\left( \dfrac{1}{2}-2\theta+2\theta^2\right)d\theta$